General solution of a Bessel-like ODE

Question

I am trying to solve the ordinary differential equation

\begin{equation*} t\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + 2\frac{\mathrm{d}\theta}{\mathrm{d}t} + c\theta = 0 \end{equation*}

[subject to $\theta(0) = \theta_{0}$ and $\theta'(0) = 0$, but I'd like to find a general solution] for $\theta(t)$, where $c>0$ is a constant, but I'm having trouble getting started. I know that the problem is soluble in terms of Bessel functions, but the equation doesn't look like any of the equations that define the Bessel functions.

I have checked several sources, namely Ince's Ordinary Differential Equations, Tenenbaum's Ordinary Differential Equations, and a couple of mathematical physics texts, but have found no help.

I don't think that Sturm-Liouville theory is readily applicable here because the equation I want to solve isn't phrased as a boundary-value problem. I appreciate any guidance, and I welcome any suggestions on how to make this question potentially interesting to a broader audience.

Edit. I have tried a power series solution whereby the equation becomes

\begin{equation*} \sum_{n=2}^{\infty}{n(n-1)a_{n}t^{n-1}} + 2\sum_{n=1}^{\infty}{na_{n}t^{n-1}} + c\sum_{n=0}^{\infty}{a_{n}t^{n}} = 0 \end{equation*}

or

\begin{equation*} \sum_{n=1}^{\infty}{n(n+1)a_{n+1}t^{n}} + 2\sum_{n=0}^{\infty}{(n+1)a_{n+1}t^{n}} + c\sum_{n=0}^{\infty}{a_{n}t^{n}} = 0 \end{equation*}

such that

\begin{equation*} 2a_{1} + ca_{0} + \sum_{n=1}^{\infty}{n(n+1)a_{n+1}t^{n}} + 2\sum_{n=1}^{\infty}{(n+1)a_{n+1}t^{n}} + c\sum_{n=1}^{\infty}{a_{n}t^{n}} = 0. \end{equation*}

Therefore, the coefficients obey

\begin{equation*} 2a_{1} + ca_{0} = 0 \hspace{1pc}\mbox{ and }\hspace{1pc}(n+1)(n+2)a_{n+1} + ca_{n} = 0. \end{equation*}

The problem I have now is that this only leaves room for a single undetermined coefficient, $a_{0}$, whereas the problem (as a second-order ODE) should have two.

Answer 1

\begin{equation*} t\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + 2\frac{\mathrm{d}\theta}{\mathrm{d}t} + c\theta = 0 \end{equation*} Change of variable $$x=2\sqrt{ct} \quad\implies\quad \frac{d\theta}{dt}=\frac{2c}{x}\frac{d\theta}{dx}\quad\implies\quad \frac{d^2\theta}{dt^2}=\frac{4c^2}{x^2}\frac{d^2\theta}{dx^2}-\frac{4c^2}{x^3}\frac{d\theta}{dx}$$ After simplification : $$\frac{d^2\theta}{dx^2} + \frac{3}{x}\frac{d\theta}{dx} + \theta = 0$$ This is a Bessel ODE which general solution is : $$\theta(x)=c_1\frac{J_1(x)}{x}+c_2\frac{Y_1(x)}{x}$$ $J_1(x)$ and $Y_1(x)$ are the Bessel functions of first and second kind respectively. $$\theta(t)=C_1\frac{J_1(2\sqrt{ct})}{\sqrt{t}}+C_2\frac{Y_1(2\sqrt{ct})}{\sqrt{t}}$$

Note : In your calculus the kind of series that you chose is convenient for the Bessel function of the first kind but not convenient for the Bessel function of the second kind. That is why no second arbirary constant $C_2$ is found.

Answer 2

The form of the equation suggests the ansatz $$t = \alpha s^\beta, \theta = t^\gamma u = \alpha^\gamma s^{\beta \gamma} u,$$ which transforms the equation to $$s^2 u''(s) + (2 \beta \gamma + \beta + 1) s u'(s) + \beta^2 (\alpha c s^\beta + \gamma^2 + \gamma) u(s) = 0 .$$ Taking $\alpha = -\frac1{4 c}, \beta = 2, \gamma = -\frac12$ transforms the equation to the Bessel equation with parameter $1$: $$s^2 u''(s) + s u'(s) + (s^2 - 1) u(s) = 0 .$$

The general solution is $$u(s) = a J_1(s) + b Y_1(s),$$ where $J_\bullet, Y_\bullet$ respectively denote the Bessel functions of the first and second kinds. In terms of the original variables, we have $$\boxed{\theta(t) = \frac1{\sqrt t} (a J_1(2 \sqrt{ct}) + b Y_1(2 \sqrt{ct}))} .$$